Determine which part/s of the following IUPAC name made them incorrect? 20 poim Locant Parent length Prefix None 4- Bromocyclopentanone 3-Hexanal 4-Ethylpentanedioic acid Isopropylbenzoate 4-Propylheptanoic acid N.N-Diethyl ethane Butylthiomethane 3-Phenylbutanamide 1-Bromopropane nitrile 4-(Ethylthio)-2-Methyl- 2-pentene O O O 0 0 O O O 0 0 O O O
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- A 5.00-mL aliquot of a solution that contains 3.05 ppm Ni2+ is treated with an appropriate excess of 2,3-quinoxalinedithiol and diluted to 50.0 mL. The molar absorptivity of a Ni2+- 2,3-quinoxalinedithiol solution at 510 nm is 5520 L mol-1 cm-1. What is the absorbance of the above diluted Ni2+- 2,3-quinoxalinedithiol solution at 510 nm in a 1.00-cm cell?In the following cases rearrange the compounds as directed : (Delhi 2010)(i) In an increasing order of basic strength :C6H5NH2, C6H5 N(CH3)2, (C2H5)2NH and CH3NH2(ii) In a decreasing order of basic strength :Aniline, p-nitroaniline and p-toluidine(iii) In an increasing order of pKb values :C2H5NH2, C6H5 NHCH3, (C2H5)2NH and C6H5NH2An unkown solvent has a formula of C4H8O and molecular weight of 72.11 g/mol. Determine the ff: a.) index of hydrohgen deficiency b.) all possible functional group present base on IHD
- Compounds Mass 3-nitrophthalic acid used 200 mg = 0.2g 8% aqueous hydrazine used 0.4mL 3-nitrophthalhydrazide obtained 130 mg = 0.13 g sodium hydrosulfite dihydrate 0.6 g luminol obtained 70 mg = 0.07 g compute yield for nitrophthalhydrazide in the first step (assume nitrophthalic acid is limiting reagent) compute yield for luminol in the second step (using nitrophthalhydrazide as limiting reagent) compute yield for the overall reaction1) A polyester made from sebacic acid (CAS# 111-20-6) and 1,6-hexanediol (CAS# 629-11-8) has the molecular weight fractionation listed below. Calculate an appropriate average molecular weight for this polyester. Source: Batzer, H., Macromolecular Chemistry and Physics 5SHOW COMPLETE SOLUTIONS Q3. Which isolate has A260/280 ratio of above 2.0?
- Compound K, L and M are three isomers with the molecular formula C5H10O.Compound K cannot be oxidized, while compound M and L can be oxidized.Oxidation of compound L with hot acidified potassium permanganate,KMnO4 solution yields 2-methylbutanoic acid. When treated with Iodoformreagent, yellow precipitation only occurs in compound K. Fehling test onlyyielded positive results for compound L and negative for compound M andK. Compound M is then reacted with hydrogen chloride, HCl produceschlorocyclopentaneCompound K, L and M are three isomers with the molecular formula C5H10O.Compound K cannot be oxidized, while compound M and L can be oxidized.Oxidation of compound L with hot acidified potassium permanganate,KMnO4 solution yields 2-methylbutanoic acid. When treated with Iodoformreagent, yellow precipitation only occurs in compound K. Fehling test onlyyielded positive results for compound L and negative for compound M andK. Compound M is then reacted with hydrogen chloride, HCl produceschlorocyclopentane (i) Draw the structural formula of compounds K, L and M.Compound K, L and M are three isomers with the molecular formula C5H10O.Compound K cannot be oxidized, while compound M and L can be oxidized.Oxidation of compound L with hot acidified potassium permanganate,KMnO4 solution yields 2-methylbutanoic acid. When treated with Iodoformreagent, yellow precipitation only occurs in compound K. Fehling test onlyyielded positive results for compound L and negative for compound M andK. Compound M is then reacted with hydrogen chloride, HCl produceschlorocyclopentane. (ii) Name the type of chemical reaction of compound M when reacted withhydrogen chloride, HCl .
- Calculate the concentration of atmospheric carbon compounds in ppbv using the data in the table below: Compound Concentration (µg/m3) Toluene 980 m,p-Xylene 910 o-Xylene 510 Benzene 370 Ethylbenzene 310 1,3,5-Trimethylbenzene 230 1-Ethyl-4-methylbenzene 200 Hexane 150 Heptane 130 1-Erthyl-2-methylbenzene 1201.Which of the following correlations about the analyses of iodine and saponificationnumbers is/are incorrect?a. Alcoholic KOH: dissolve KOH to lipid sampleb. Hanus reagent: react with saturated lipid bondc. Reflux set-up: hasten reactiond. Blank titration: determine total mol of KOH2. A triacylglycerol (Sample X) was subjected to saponification analysis. 1.1 grams of thesample was refluxed with 5% alcoholic KOH and the refluxed solution was titrated with 17.2 mLof 0.44 M HCl titrant. The blank solution was titrated with 25.6 mL of the titrant.i. What is the saponification number of the sample? ii. Which of the following could be the identity of the sample? Show solutions pls a. Tripalmitin (MW: 807.3)b. Trilaurin (MW: 639.8)c. Triolein (MW: 885:4)d. Tristearin (MW: 891.48)Multistep synthesis of Lidocaine Lab 1) How many grams of starting material do you need? It is given in moles below but how many grams? 2) How to calculate theoretical yield? Part One Method A. Preparation of alpha-Chloro-2,6-dimethylacetanilideCombine 0.033 mol of 2,6-dimethylaniline with 25 mL of glacial acetic acid (or ethyl acetateor THF) and 0.033 mol of alpha-chloroacetyl chloride, in that order, in an appropriately sized Erlenmeyer flask. With the aid of a hot water bath, warm the solution to 40-50°C, remove the flask from the bath, and add a solution of 5 g of sodium acetate trihydrate dissolved in 100 mL of water.